A given wave has a wavelength of 5.0 m and a frequency of 2.0 Hz. How fast

To give an idea of sensitivity of the platypus's electric sense, how far from a 80nC n C point charge does the field have this m

SVETLANKA909090 [29]

The question is incomplete. The complete question is :

A platypus foraging for prey can detect an electric field as small as 0.002 N/C.

-To give an idea of sensitivity of the platypus's electric sense, how far from a +80nC point charge does the field have this magnitude?

Solution :

Given electric field, E = 0.002 N/C

Charge, Q = + 80 nC

$\therefore E = \frac{kQ}{R^2}$

or $R^2=\frac{kQ}{E}$

$R^2=\frac{9\times 10^9 \times 80 \times 10^{-9}}{0.002}$

R = 600 m

This is the distance of the charge from the point of observations.

5 0

3 years ago

A motorcycle is moving at 30 m/s when the rider applies the brakes, giving the motorcycle a constant deceleration. During the 3.

shutvik [7]

Answer:

The motorcycle travelled 69.73 m during these 3.1 s.

Explanation:

In order to calculate the distance that the motorcycle travelled we first need to obtain the acceleration rate that was used to brake the vehicle. We do that by using the following formula:

a = (V_final - V_initial)/(t) = (15 - 30)/(3.1) = -4.84 m/s^2

The distance is given by the following formula:

S = (V_final^2 - V_initial^2)/(2*a)

S = (15^2 - 30^2)/[2*(-4.84)] = (225 - 900)/(-9.68) = -675/(-9.68) = 69.73 m

The motorcycle travelled 69.73 m during these 3.1 s.

5 0

3 years ago

Using numbers to describe kinematic quantities is called what description?

MissTica

sorry dont know this

4 0

3 years ago

A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm

Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

$F = - k * \Delta x (1)$

where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
Solving for k:

$k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)$

b)

Assuming no friction present, total mechanical energy mus keep constant.
When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

$U = \frac{1}{2}* k* (\Delta x)^{2} (3)$

Replacing k and Δx by their values, we get:

$U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)$

c)

When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
So, we can write the following expression:

$\frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2} = \frac{1}{2}*k*\Delta x^{2} (5)$